Stop audio from playing using sound()

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YT
YT 2019년 2월 12일
댓글: Nguyen Nam 2024년 4월 13일 3:04
Matlab Online (ver R2018b)
I'm playing some audio using sound(y,Fs) but I can't find a way to stop the audio from playing. You can guess that it gets even worse when you (accidentally) execute the command twice (yes it plays the music over eachother instead of replacing it). Refreshing the page is the only option for now.
p.s.
I know you can use audioplayer(y,Fs) for offline versions but it won't work online.
  댓글 수: 1
Jan
Jan 2022년 3월 11일
The sound() command does call audioplayer() internally in R2018b. So I cannot see a reason to prefer sound() instead or using audioplayer() directly.

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채택된 답변

Walter Roberson
Walter Roberson 2019년 2월 12일
sound() can only be interrupted by killing matlab .
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Walter Roberson
Walter Roberson 2021년 8월 8일
sound() keeps a persistent variable that is a vector of audioplayer objects. Each time you use sound(), it looks at its existing vector and destroys the ones that have stopped playing. Then it creates an audioplayer object and adds it to the end of the list; it does not record any identification information such as a unique tag when it does that.
In order to stop a specific sound() you would need to get access to the list of audioplayer objects that is stored inside the function, at the same time ignoring any audioplayer objects that were created outside the function (because those ones cannot be the right sound.)
Unfortunately, audioplayer does not derive from handle class, and audioplayer class does not keep track of audioplayer objects, and mathworks does not provide any documented method of accessing the persistent variables inside a function. It is not obvious that the objects can be examined at all.
There just might be a way using the kinds of ideas outlined in https://www.mathworks.com/matlabcentral/answers/474974-how-can-i-find-all-object-of-a-certain-handle-class#answer_386321 but note that is talking about handle classes and audioplayer is not a handle class.
If, hypothetically, you were able to locate the audioplayer objects, then you run into the problem that the objects do not have any tags or identifiers. If you knew the size() of the data that was being played, you might be able to reduce the search by removing from consideration the ones that do not have the right size, but you would pretty much have to compare passed-in sound to the data stored in the audioplayer object in order to determine whether you had the right object.
It seems to me that it would be a lot easier to just use audioplayer() yourself and keep track of the objects.
Jan
Jan 2022년 3월 11일
@Walter Roberson: Do you agree to accept Joe V's answer, because it is working?

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추가 답변 (2개)

Joe V
Joe V 2019년 6월 27일
  댓글 수: 4
Marcel-Dennis Boerzel
Marcel-Dennis Boerzel 2022년 2월 22일
works. Thanks!!!
Nguyen Nam
Nguyen Nam 2024년 4월 13일 3:04
Thanks.

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MUHAMMAD HAIDIEL SHAFIS BIN MAZELAN
lol u guys can also use this too...
[a, fs] = audioread('yourfilehere.mp3');
plot (a(1:100000,1))
fs =1.0*fs; %%We can adjust our speed of our song by multiplying our fs with any desired numbers.
sound (a,fs);
y = input ('Press 1 to stop the music: ');
while y~=1
fprintf ('You didn''t enter 1 ');
fprintf ('\n');
end
clear sound;
  댓글 수: 2
Walter Roberson
Walter Roberson 2021년 1월 6일
If the person does not enter 1, then your while is an infinite loop.
Rik
Rik 2021년 1월 6일
Also, the relevant code is equivalent to the other answer. What does this one add?

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